Question

The magnitude of electric force on $2\mu c$ change placed at the centre $O$ of two equilateral triangle each of side $10cm$, as shown in figure is $P$. If charge $A,B,C,D,E$ & $F$ are $2\mu c,2\mu c,2\mu c,-2\mu c,-2\mu c,-2\mu c$ respectively, then $P$ is:

• A. $21.6N$
• B. $64.8N$
• C. $0$
• D. $43.2N$

Answer 1

The correct option is C $0$

$q_0=2\mu C$

$q_A=2\mu C$

$q_B=2\mu C$

$q_C=2\mu C$

$q_D=-2\mu C$

$q_E=-2\mu C$

$q_F=-2\mu C$

Side of the triangle $=10cm$

$r=\frac{5}{2}\sqrt{5}$ [Visit from O to any vertex]

$F_{AO}=\frac{1}{4\pi {\epsilon }_0}.\frac{q_A{q}_0}{r^2}=\frac{9\times {10}^9\times 2\times 2\times {10}^{-12}}{\frac{25}{4}\left(5\right)}=0.001152N$

$F_{BO}=\frac{1}{4\pi {\epsilon }_0}.\frac{q_B{q}_0}{r^2}=\frac{9\times {10}^9\times 2\times 2\times {10}^{-12}}{\frac{25}{4}\left(5\right)}=0.001152N$

$F_C=\frac{1}{4\pi {\epsilon }_0}.\frac{q_C{q}_0}{r^2}=\frac{9\times {10}^9\times 2\times 2\times {10}^{-12}}{\frac{25}{4}\left(5\right)}=0.001152N$

$F_D=\frac{1}{4\pi {\epsilon }_0}.\frac{q_D{q}_0}{r^2}=-\frac{9\times {10}^9\times 2\times 2\times {10}^{-12}}{\frac{25}{4}\left(5\right)}=-0.001152N$

$F_E=\frac{1}{4\pi {\epsilon }_0}.\frac{q_E{q}_0}{r^2}=-\frac{9\times {10}^9\times 2\times 2\times {10}^{-12}}{\frac{25}{4}\left(5\right)}=-0.001152N$

$F_{net}=F_{AO}+F_{BO}+F_C+F_D+F_E+F_F=ON$